If the solute is properly distributed in the given volume, there are 2.642 g of (NH4)2SO4 per 10 mL. For the new solution, divide the 2.642 g by the molar mass of the compound. The answer is 0.02 moles. Then, divide this by the new volume, 50 mL or 0.05 L. The concentration of the new solution is 0.4 M.
Aluminum oxide produced : = 79.152 g
<h3>Further explanation</h3>
Given
46.5g of Al
165.37g of MnO
Required
Aluminum oxide produced
Solution
Reaction
2 Al (s) + 3 MnO (s) → 3 Mn (s) + Al₂O₃ (s)
mol = mass : Ar
mol = 46.5 : 27
mol = 1.722
mol = 165.37 : 71
mol = 2.329
mol : coefficient ratio Al : MnO = 1.722/2 : 2.329/3 = 0.861 : 0.776
MnO as a limiting reactant(smaller ratio)
So mol Al₂O₃ based on MnO as a limiting reactant
From equation , mol Al₂O₃ :
= 1/3 x mol MnO
= 1/3 x 2.329
= 0.776
Mass Al₂O₃ (MW=102 g/mol) :
= 0.776 x 102
= 79.152 g
Answer: 0.004 moles
Explanation:
Amount of substance= mass/ molar mass
Mass= 250mg or 0.25g
Molar mass= 58.44gmol-1
Amount= 0.25/58.44
= 0.004 moles
Glucose and a plants and ur welcome
Answer:
V₂ = 0.62 L
Explanation:
Given data:
Initial volume = 2.4 L
Initial temperature = 25°C
Final temperature = -196°C
Final volume = ?
Solution:
Initial temperature = 25°C (25+273 = 298 K)
Final temperature = -196°C ( -196+273 = 77 K)
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 2.4 L × 77 K / 298 k
V₂ = 184.8 L.K / 298 K
V₂ = 0.62 L
